Are you considering $d$ to be an arbitrary positive integer?
Then the answer is yes. In fact, one can show the following is true:
For the homogeneous equation with real coefficients:
$$\tag{1}\def\sss{}
a_{\sss n} y^{\sss(\!n\!)} +a_{\sss n\!-\!1} y^{\sss(\!n\!-\!1\!
)} +\cdots+a_{\sss1} y' +a_{\sss0} y = 0,\quad a_n\ne0
$$
The characteristic polynomial (c.p.) is
$$
\tag{2}a_{\sss n}x^n+a_{\sss n\!-\!1}x^{n-1} +\cdots+a_{\sss1}x +a_{\sss0} .
$$
Find the roots and their multiplicities of (2).
If $c$ is a real root of (2) with multiplicity $k $, then $k$-independent solutions of (1)
are
$$
e^{ct}, xe^{ct}, x^2 e^{ct}, \ldots, x^{k-1}e^{ct}
$$
If $a+bi$ is a complex root of (2) with multiplicity $k$, then $2k$-independent solutions of (1)
are
$$
e^{at}\sin (bt), x e^{at}\sin (bt), \ldots, x^{k-1} e^{at}\sin (bt)
$$
$$
e^{at}\cos (bt), xe^{at}\cos (bt) , \ldots, x^{k-1}e^{at}\cos (bt)
$$
Moreover, the set of all solutions found from the above will be independent (for instance, one can compute the Wronskian) and there will be $n$ of them (this follows from the Fundamental Theorem of Algebra).
The general solution to (1) is
$$y_c=c_{\sss 1}y_{\sss1}+c_{\sss2}y_{\sss2}+\cdots+c_{\sss n}y_{\sss n}$$
where $y_1$, $y_2$, $\ldots\,$, $y_n$ are the $n$-solutions found above.
Let $y_1,y_2$ be solutions for the equation
$$y''+ay'+by=0.$$
Then
$$y_1''+ay_1'+by_1=0$$
$$y_2''+ay_2'+by_2=0$$
Add the first and the second equation we have
$$(y_1+y_2)''+a(y_1+y_2)'+b(y_1+y_2)=0.$$
Then $y_1+y_2$ is solution too.
Let $\lambda$ be a real number
$$\lambda y_1''+a\lambda y_1'+b\lambda y_1=\lambda(y_1''+ay_1'+by_1)=0.$$
Then $\lambda y_1$ is solution too.
Best Answer
The solution space of any homogeneous linear differential equation of order $m$ (on an interval in which the Existence and Uniqueness Theorem applies) has dimension $m$. One way to see this is that you can parametrize the solutions by the values of $y, y', \ldots, y^{(m-1)}$ at a particular $x$ in the interval: for any such initial condition there is exactly one solution to the differential equation that satisfies that initial condition.
In this case it's a second order equation, so the dimension is indeed $2$.